Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $y = \dfrac{-8p + 56}{p^2 - 12p + 36} \div \dfrac{p - 7}{3p^2 - 18p} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{-8p + 56}{p^2 - 12p + 36} \times \dfrac{3p^2 - 18p}{p - 7} $ First factor the quadratic. $y = \dfrac{-8p + 56}{(p - 6)(p - 6)} \times \dfrac{3p^2 - 18p}{p - 7} $ Then factor out any other terms. $y = \dfrac{-8(p - 7)}{(p - 6)(p - 6)} \times \dfrac{3p(p - 6)}{p - 7} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ -8(p - 7) \times 3p(p - 6) } { (p - 6)(p - 6) \times (p - 7) } $ $y = \dfrac{ -24p(p - 7)(p - 6)}{ (p - 6)(p - 6)(p - 7)} $ Notice that $(p - 7)$ and $(p - 6)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -24p(p - 7)\cancel{(p - 6)}}{ \cancel{(p - 6)}(p - 6)(p - 7)} $ We are dividing by $p - 6$ , so $p - 6 \neq 0$ Therefore, $p \neq 6$ $y = \dfrac{ -24p\cancel{(p - 7)}\cancel{(p - 6)}}{ \cancel{(p - 6)}(p - 6)\cancel{(p - 7)}} $ We are dividing by $p - 7$ , so $p - 7 \neq 0$ Therefore, $p \neq 7$ $y = \dfrac{-24p}{p - 6} ; \space p \neq 6 ; \space p \neq 7 $